276 Mechanics of Materials i.e. TADA TEDB JA JB TA DBJA TB DAJB (2) But the strain energy of B=1.25 x strain energy of A. since U= TL 2GJ TiL then 2GB1.25 TiL or T JA GJA T扇-1.25JB Therefore substituting from(2), D后JB D1.25J4 2[D哈-d] Dt-d进 D 125品时 1.25D4 p%-5D D哈-dt =D8-(D后-10×10-3y 1.25×10×10-3 12.5×103D后=D%-D8+20×10-3D2-100×10-6 7.5×10-3×D2=100×10-6 100×10-6 D%=73x10可=133×10- Dg=115.47mm =D3-0时-1得3-0-3 103-103=109 da=57.74mm The internal and external diameters of the hollow tube are therefore 57.7 mm and 115.5 mm respectively. Example 11.3 (a)What will be the instantaneous stress and elongation of a 25 mm diameter bar,2.6m long,suspended vertically,if a mass of 10kg falls through a height of 300 mm on to a collar which is rigidly attached to the bottom end of the bar? Take g 10m/s2.276 Mechanics of Materials i.e. But the strain energy of B = 1.25 x strain energy of A. then T’, JA -- T;L T; L - 1.25- or -- E&- ~GJA Ti 1.25JB Therefore substituting from (2), D; JB D’, 1.255, - =--- - 0;- (0:- 10 x 10-3)2 - 1.25 x io 10-3 12.5 x lop3 0% = D; - D;+ 20 x Di- 100 x .. 7.5 x 10-3 x ~zg= 100 10-6 100 x 10-6 Dzg= = 13.3 x 10-3 7.5 x 10-3 DB = 115.47 mm 13.3 10 3.3 di= Di-D’, = =- io3 io3 103 .. dB = 57.74 mm 115.5 mm respectively. The internal and external diameters of the hollow tube are therefore 57.7mm and Example 11.3 (a) What will be the instantaneous stress and elongation of a 25 mm diameter bar, 2.6 m long, suspended vertically, if a mass of 10 kg falls through a height of 300 mm on to a collar which is rigidly attached to the bottom end of the bar? Take g = 10m/s2